TRUYỀN SỐ LIỆU VÀ MẠNG Ch01 transmission media and phy layer

Data Communication and Networking Dr –Ing Vo Que Son Email: sonvq@hcmut.edu.vn Telecomm Dept Faculty of EEE DCN HCMUT Content Chapter 1: Transmission Media and PHY Layer Wired and Wireless Media Physical layer standards: RS232, RS422, RS485 Line Coding Digital modulation/demodulation Channel parameters Gaussian noise and BER Chapter 2: Data Communication Asynchronous data transmission Synchronous data transmission Channel Coding Data Compression Telecomm Dept Faculty of EEE DCN HCMUT Wired Media  Guided Media  How can signal be transmitted in wired media (cables)?  Voltage is sometimes referred to as electromotive force (EMF)  EMF is related to an electrical force, or pressure, that occurs when electrons and protons are separated Force within a atom Static Electricity neutron proton Electrostatic discharge Telecomm Dept Faculty of EEE DCN HCMUT Waves Sine Wave Square Wave • Repeat the same pattern at regular intervals • Repeat the same pattern at regular intervals • Continuous voltage •do not continuously voltage • occur naturally and change regularly over time • Repeat the flat pattern on both the top and bottom of the wave Telecomm Dept Faculty of EEE DCN HCMUT Two-wire open lines Used in short distance communication with low data rates Simple structure Data rate < 19Kbps, max distance LVT: decide the received bit is bit  VD vT) = Pr(1/0) =  vT 2 e  x2 2 dx  Assuming that the probability of transmitting bit and bit is the same: Pr(1)=Pr(0)=0.5 Telecomm Dept Faculty of EEE DCN HCMUT 135 Gaussian noise and BER  Probability of bit error (Bayes rule): Pe = Pr(1).Pr(0/1)+ Pr(0).Pr(1/0)  Because Pr(0)=Pr(1)=0.5, then: Pe = Pr(0/1)=Pr(1/0)  Question: what is the value of VT so that Pe is minimum? v T Pe =0.5(  - e 2πσ2 -(v T -A)2 -v T2 2σ2 2σ2 Telecomm Dept Faculty of EEE -e e =0 -(x-A)2 2σ2  dx+  vT  2πσ2 -x2 2σ2 e dx) (v T -A)2 =v 2T minimum  v T =A/2 DCN HCMUT 136 Gaussian noise and BER Finally, we need to determine Pr(1/0) or Pr(0/1): using Q(k) function Q(k) is a normal distribution function m=0, =1 In this case: VT=A/2(why?) VT/=A/2 so: Pe=Q(A/2) Area is Q(k) Let k= A/2: Pe=Q(k) Telecomm Dept Faculty of EEE DCN HCMUT 137 Gaussian noise and BER How to determine Q(k) Q(k) chart for k7: using the formula Q(k)  e 2 k Telecomm Dept Faculty of EEE k2  DCN HCMUT 138 Frame Error Rate  Given Pe  If a frame has n bits What is the Frame Error Rate?  Probability of k-bit error: Pk =CnkPek (1-Pe )n-k  P0 is the probability that there is no errors in the frame  Frame Error Rate (FER): Pf(error) = 1-P0 = 1-(1-Pe )n  nPe (because Pe